Answer:
There is no question. Please provide a question.
Answer:
0.75 mg
Step-by-step explanation:
From the question given above the following data were obtained:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Amount remaining (N) =.?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Number of half-lives (n) =?
n = t / t₁/₂
n = 6/6
n = 1
Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Number of half-lives (n) = 1
Amount remaining (N) =.?
N = 1/2ⁿ × N₀
N = 1/2¹ × 1.5
N = 1/2 × 1.5
N = 0.5 × 1.5
N = 0.75 mg
Thus, 0.75 mg of the sample is remaining.
25hr. 40h. 10hr
375$ 600$ x
25 . 40 . x = 1000
375 . 600 = 225000
225000÷1000
x = 225
Answer: 0.0228
Step-by-step explanation:
Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-
If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.
Let X be the random variable that represents the finish times for this event.
z score : 
Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-
Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228
Answer:
3 terms we have constant ,Y and X terms
